#!/usr/bin/python3
"""
Given a non-empty, singly linked list with head node head, return a middle node
of linked list.

If there are two middle nodes, return the second middle node.

Example 1:

Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3.  (The judge's serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
Example 2:

Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.

Note:

The number of nodes in the given list will be between 1 and 100.
"""


# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        """
        """
        l = 0
        cur = head
        while cur:
            l += 1
            cur = cur.next

        mid = l // 2 + 1
        cur_l = 0
        cur = head
        while cur:
            cur_l += 1
            if cur_l == mid:
                return cur
            cur = cur.next

        return None
